Calculus Examples

$\frac{x^{4} - 2 x^{3} + x^{2} + 6 x - 5}{x^{2} - 2 x + 1}$

Step 1

Divide using long polynomial division.

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Step 1.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x^{2}$

$2 x$

$1$

$x^{4}$

$2 x^{3}$

$x^{2}$

$6 x$

$5$

Step 1.2

Divide the highest order term in the dividend $x^{4}$ by the highest order term in divisor $x^{2}$ .

$x^{2}$

$2 x$

$1$

$x^{4}$

$2 x^{3}$

$x^{2}$

$6 x$

$5$

Step 1.3

Multiply the new quotient term by the divisor.

$x^{2}$

$2 x$

$1$

$x^{4}$

$2 x^{3}$

$x^{2}$

$6 x$

$5$

$x^{4}$

$2 x^{3}$

$x^{2}$

Step 1.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{4} - 2 x^{3} + x^{2}$

$x^{2}$

$2 x$

$1$

$x^{4}$

$2 x^{3}$

$x^{2}$

$6 x$

$5$

$x^{4}$

$2 x^{3}$

$x^{2}$

Step 1.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

$x^{2}$

$2 x$

$1$

$x^{4}$

$2 x^{3}$

$x^{2}$

$6 x$

$5$

$x^{4}$

$2 x^{3}$

$x^{2}$

$0$

Step 1.6

Pull the next terms from the original dividend down into the current dividend.

$x^{2}$

$2 x$

$1$

$x^{4}$

$2 x^{3}$

$x^{2}$

$6 x$

$5$

$x^{4}$

$2 x^{3}$

$x^{2}$

$0$

$6 x$

$5$

Step 1.7

The final answer is the quotient plus the remainder over the divisor.

$x^{2} + \frac{6 x - 5}{x^{2} - 2 x + 1}$

Step 2

Decompose the fraction and multiply through by the common denominator.

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Step 2.1

Factor using the perfect square rule.

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Step 2.1.1

Rewrite $1$ as $1^{2}$ .

$\frac{6 x - 5}{x^{2} - 2 x + 1^{2}}$

Step 2.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 2.1.3

Rewrite the polynomial.

$\frac{6 x - 5}{x^{2} - 2 \cdot x \cdot 1 + 1^{2}}$

Step 2.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

$\frac{6 x - 5}{{(x - 1)}^{2}}$

Step 2.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $A$ .

$\frac{A}{{(x - 1)}^{2}}$

Step 2.3

$\frac{A}{{(x - 1)}^{2}} + \frac{B}{x - 1}$

Step 2.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is ${(x - 1)}^{2}$ .

$\frac{(6 x - 5) {(x - 1)}^{2}}{{(x - 1)}^{2}} = \frac{(A) {(x - 1)}^{2}}{{(x - 1)}^{2}} + \frac{(B) {(x - 1)}^{2}}{x - 1}$

Step 2.5

Cancel the common factor of ${(x - 1)}^{2}$ .

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Step 2.5.1

Cancel the common factor.

$\frac{(6 x - 5) {(x - 1)}^{2}}{{(x - 1)}^{2}} = \frac{(A) {(x - 1)}^{2}}{{(x - 1)}^{2}} + \frac{(B) {(x - 1)}^{2}}{x - 1}$

Step 2.5.2

Divide $6 x - 5$ by $1$ .

$6 x - 5 = \frac{(A) {(x - 1)}^{2}}{{(x - 1)}^{2}} + \frac{(B) {(x - 1)}^{2}}{x - 1}$

Step 2.6

Simplify each term.

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Step 2.6.1

Cancel the common factor of ${(x - 1)}^{2}$ .

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Step 2.6.1.1

Cancel the common factor.

$6 x - 5 = \frac{A {(x - 1)}^{2}}{{(x - 1)}^{2}} + \frac{(B) {(x - 1)}^{2}}{x - 1}$

Step 2.6.1.2

Divide $A$ by $1$ .

$6 x - 5 = A + \frac{(B) {(x - 1)}^{2}}{x - 1}$

Step 2.6.2

Cancel the common factor of ${(x - 1)}^{2}$ and $x - 1$ .

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Step 2.6.2.1

Factor $x - 1$ out of $(B) {(x - 1)}^{2}$ .

$6 x - 5 = A + \frac{(x - 1) (B (x - 1))}{x - 1}$

Step 2.6.2.2

Cancel the common factors.

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Step 2.6.2.2.1

Multiply by $1$ .

$6 x - 5 = A + \frac{(x - 1) (B (x - 1))}{(x - 1) \cdot 1}$

Step 2.6.2.2.2

Cancel the common factor.

$6 x - 5 = A + \frac{(x - 1) (B (x - 1))}{(x - 1) \cdot 1}$

Step 2.6.2.2.3

Rewrite the expression.

$6 x - 5 = A + \frac{B (x - 1)}{1}$

Step 2.6.2.2.4

Divide $B (x - 1)$ by $1$ .

$6 x - 5 = A + B (x - 1)$

Step 2.6.3

Apply the distributive property.

$6 x - 5 = A + B x + B \cdot - 1$

Step 2.6.4

Move $- 1$ to the left of $B$ .

$6 x - 5 = A + B x - 1 \cdot B$

Step 2.6.5

Rewrite $- 1 B$ as $- B$ .

$6 x - 5 = A + B x - B$

Step 2.7

Reorder $A$ and $B x$ .

$6 x - 5 = B x + A - B$

Step 3

Create equations for the partial fraction variables and use them to set up a system of equations.

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Step 3.1

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$6 = B$

Step 3.2

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$- 5 = A - 1 B$

Step 3.3

Set up the system of equations to find the coefficients of the partial fractions.

$6 = B$

$- 5 = A - 1 B$

$6 = B$

$- 5 = A - 1 B$

Step 4

Solve the system of equations.

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Step 4.1

Rewrite the equation as $B = 6$ .

$B = 6$

$- 5 = A - 1 B$

Step 4.2

Replace all occurrences of $B$ with $6$ in each equation.

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Step 4.2.1

Replace all occurrences of $B$ in $- 5 = A - 1 B$ with $6$ .

$- 5 = A - 1 \cdot 6$

$B = 6$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Multiply $- 1$ by $6$ .

$- 5 = A - 6$

$B = 6$

$- 5 = A - 6$

$B = 6$

$- 5 = A - 6$

$B = 6$

Step 4.3

Solve for $A$ in $- 5 = A - 6$ .

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Step 4.3.1

Rewrite the equation as $A - 6 = - 5$ .

$A - 6 = - 5$

$B = 6$

Step 4.3.2

Move all terms not containing $A$ to the right side of the equation.

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Step 4.3.2.1

Add $6$ to both sides of the equation.

$A = - 5 + 6$

$B = 6$

Step 4.3.2.2

Add $- 5$ and $6$ .

$A = 1$

$B = 6$

$A = 1$

$B = 6$

$A = 1$

$B = 6$

Step 4.4

Solve the system of equations.

$A = 1 B = 6$

Step 4.5

List all of the solutions.

$A = 1, B = 6$

Step 5

Replace each of the partial fraction coefficients in $x^{2} + \frac{A}{{(x - 1)}^{2}} + \frac{B}{x - 1}$ with the values found for $A$ and $B$ .

$x^{2} + \frac{1}{{(x - 1)}^{2}} + \frac{6}{x - 1}$